Note: See update below.
Consider the Majorana Lagrangian
$$\mathcal{L}=-\psi ^{\mathrm{T}}\mathrm{i}% \gamma ^{0}\left( \gamma ^{\rho }\partial _{\rho }+m\right) \psi ,\tag{1}$$
where $% \psi \in \mathbb{R}^{4}$ is Grassmann-valued, and where $\gamma ^{\rho }\in \mathrm{M}_{4}\left( \mathbb{R}\right) $ is the irreducible, real-valued representation of $\mathrm{Cliff}\left( 3,1\right) $. The generalized coordinates $\psi ^{a}=\psi ^{a}\left( \mathbf{x}\right) $ and their conjugate momenta $\pi _{a}\left( \mathbf{x}\right) \equiv \pi _{a}^{L}\left( \mathbf{x% }\right) =-\pi _{a}^{R}\left( \mathbf{x}\right) $, where
\begin{eqnarray*} \pi _{a}^{L} &\equiv &\frac{\partial _{L}\mathcal{L}}{\partial \dot{\psi}^{a}% }\equiv \frac{\overrightarrow{\partial }}{\partial \dot{\psi}^{a}}\mathcal{L}% =+\mathrm{i}\left( \psi ^{\mathrm{T}}\right) _{a}=+\mathrm{i}\psi _{a}, \\ \pi _{a}^{R} &\equiv &\frac{\partial _{R}\mathcal{L}}{\partial \dot{\psi}^{a}% }\equiv \mathcal{L}\frac{\overleftarrow{\partial }}{\partial \dot{\psi}^{a}}% =-\mathrm{i}\left( \psi ^{\mathrm{T}}\right) _{a}=-\mathrm{i}\psi _{a},\tag{2} \end{eqnarray*}
following answer by Qmechanic, are obviously not independent, but constrained by $$ 0=\chi _{a \mathbf{x}} \equiv \psi _{a}\left( \mathbf{x}\right) +\mathrm{i}\pi _{a}\left( \mathbf{x}\right) .\tag{3} $$ The Poisson brackets of these constraints with themselves are given by
\begin{eqnarray*} C_{a \mathbf{x},b \mathbf{y}} &\equiv &\left\{ \chi _{a \mathbf{x}},\chi _{b \mathbf{y}} \right\} _{\text{P}} \\ &=&\mathrm{i}\left\{ \psi _{a}\left( \mathbf{x}\right) ,\pi _{b}\left( \mathbf{y}\right) \right\} _{\text{P}}+\mathrm{i}\left\{ \pi _{a}\left( \mathbf{x}\right) ,\psi _{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\ &=&-2\mathrm{i}\delta _{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}-% \mathbf{y}\right) ,\tag{4} \end{eqnarray*}
using
\begin{eqnarray*} \left\{ F,G\right\} _{\text{P}} &=&\int d^{3}x\left[ \frac{\delta _{R}F}{ \delta \psi ^{a}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \pi _{a}^{R}\left( \mathbf{x}\right) }-\frac{\delta _{R}F}{\delta \pi _{a}^{L}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \psi ^{a}\left( \mathbf{x}\right) }\right] \\ &\equiv &-\int d^{3}x\left[ \frac{\delta _{R}F}{\delta \psi ^{a}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \pi _{a}\left( \mathbf{x} \right) }+\frac{\delta _{R}F}{\delta \pi _{a}\left( \mathbf{x}\right) }\frac{ \delta _{L}G}{\delta \psi ^{a}\left( \mathbf{x}\right) }\right] ,\tag{5} \end{eqnarray*}
following again answer by Qmechanic. The matrix $C_{a \mathbf{x},b \mathbf{y}}$ being invertible,
$$ \left( C^{-1}\right) _{a\mathbf{x},b\mathbf{y}}=\frac{\mathrm{i}}{2}\delta _{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{y}\right),\tag{6} $$
implies, in the terminology of Dirac, that these constraints are second class. The Dirac bracket is thus given by
\begin{eqnarray*} \left\{ F,G\right\} _{\text{D}} &\equiv &\left\{ F,G\right\} _{\text{P} }-\int d^{3}x\int d^{3}y\left\{ F,\chi ^{a\mathbf{x}}\right\} _{\text{P} }\left( C^{-1}\right) _{a\mathbf{x},b\mathbf{y}}\left\{ \chi ^{b\mathbf{y} },G\right\} _{\text{P}} \\ &=&\left\{ F,G\right\} _{\text{P}}-\frac{\mathrm{i}}{2}\int d^{3}x\left\{ F,\chi ^{a\mathbf{x}}\right\} _{\text{P}}\left\{ \chi _{a\mathbf{x} },G\right\} _{\text{P}},\tag{7} \end{eqnarray*}
so that
\begin{eqnarray*} \left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{D}} &=&\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}}-\frac{\mathrm{i}}{2}\int d^{3}z\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\chi ^{c\mathbf{z}}\right\} _{\text{P}}\left\{ \chi _{c\mathbf{z}},\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\ &=&\frac{\mathrm{i}}{2}\int d^{3}z\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\pi ^{c}\left( \mathbf{z}\right) \right\} _{\text{P}}\left\{ \pi _{c}\left( \mathbf{z}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\ &=&\frac{\mathrm{i}}{2}\int d^{3}z\delta ^{ac}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{z}\right) \delta _{c}^{b}\delta ^{\left( 3\right) }\left( \mathbf{z}-\mathbf{y}\right) \\ &=&\frac{\mathrm{i}}{2}\delta ^{ab}\delta ^{\left( 3\right) }\left( \mathbf{x }-\mathbf{y}\right) .\tag{8} \end{eqnarray*}
Following the 'quantum bracket = $\mathrm{i} \times$ Poisson bracket'-rule, the quantum bracket is thus supposedly given by
$$ \left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} =-\frac{1}{2}\delta ^{ab}\delta ^{\left( 3\right) }\left( \mathbf{x} -\mathbf{y}\right) .\tag{9} $$
Can that really be correct? The minus sign looks wrong. And what about the factor of 1/2? But before turning attention to these details, perhaps I should start by asking whether I have made some conceptual errors above. Have I simply misunderstood how to go about quantizing constrained systems?
Update: Now the penny finally dropped. Almost embarrassingly, the wrong sign boils down to not having taken into account the minus sign in $\gamma _{0}^{2}=-1_{4}$ in the calculation of the conjugate momenta. Taking proper care of this, the sign issue evaporates. (Note that in order to preserve the history of this posting, the material above has not been edited accordingly.) Thus to me only remains now the puzzlement over the factor of 1/2, see my comment below.
